3
$\begingroup$

I'm reading an example for abstract algebra and I would like to clarify some terms. The example is showing the number of homomorphisms from $\mathbb Z $ to $S_3$.

Take a homomorphism $f$ from $\mathbb Z$ to $S_3$. Since $1$ is a generator of $\mathbb Z$, $f$ is completely determined by the image of $1$. Since $1$ has infinite order, there are no restrictions on the image of $1$. Thus each element of $S_3$ is a possible image for $1$. There are $6$ homomorphisms from $\mathbb Z$ to $S_3$.

What does "completely determined by the image of $1$" mean? I thought about it a little bit, and it seems that since $1$ is a generator of $\mathbb Z$, any element in $\mathbb Z$ can be generated from $1^k$. When it says image of $1$, does it mean $f(1^k)$, where $k$ is an integer? In general, what does the "image" of an element under homomorphism mean? I thought for an element $g$ in group $G$, and for a homomorphism $h$, the image of $g$ under $h$ is $h(g)$. How can that determine the entire homomorphism?

Thanks for your help.

$\endgroup$
  • $\begingroup$ As $\mathbb Z$ is written additively, it qwould be less confusing to write $k\cdot 1$ instead of $1^k$ (after all, it is supposed to mean $\underbrace{1+1+\ldots+1}_{k}$ $\endgroup$ – Hagen von Eitzen Oct 19 '14 at 20:32
  • $\begingroup$ Thanks Hagen. How did you fix up the formatting? Would like to learn for future posts. $\endgroup$ – jstnchng Oct 19 '14 at 20:52
  • $\begingroup$ @JC1397 You can use LaTeX both inline with $ and as a paragraph with $$ delimiters. Read the editing help for details. $\endgroup$ – Lukas Oct 19 '14 at 21:42
3
$\begingroup$

You are right. The image of $g\in A$ under a homomorphism $h\colon A\to B$ is the element $h(g)\in B$ to which $g$ is mapped.

That $f\colon \mathbb Z\to S_3$ is completely determined by the image of $1$ means that if you have two homomorphisms $f_,g\colon \mathbb Z\to S_3$ and the image of $1$ is the same for both (i.e. $f(1)=g(1)$) then the homomorphisms are the same (i.e. $f(x)=g(x)$ for all $x\in\mathbb Z$). More generally, if $A$ is a cyclic group with generator $a$ and $B$ is any group, then a homomorphism $f\colon A\to B$ is uniquely determined by the image of $a$. This follows from the homomorphism properties of $f$ and $g$ because $f(a)=g(a)$ gives us $f(a^{-1})=g(a^{-1})$ and then by induction $f(a^{\pm n})=g(a^{\pm n})$. Since $a$ generates $A$, this covers all elements of $A$. Even more generally, the corresponding statement holds also for groups with more generators.

$\endgroup$
4
$\begingroup$

It's a general property of algebraic structures, but let's limit ourselves to groups.

For any subset $S$ of $G$ there is the least subgroup $\langle S\rangle$ of $G$ containing $S$. It can be described as the set of all elements of the form $$ s_1^{\varepsilon_1}s_2^{\varepsilon_2}\dots s_n^{\varepsilon_n} $$ where $n$ is an arbitrary integer, $s_1,s_2,\dots,s_n\in S$ and $\varepsilon_k=\pm1$ for $k=1,2,\dots,n$.

The subset $S$ is said to be a generating set for $G$ if $\langle S\rangle=G$.

If $f\colon G\to G'$ and $g\colon G\to G'$ are homomorphisms, $S$ is a generating set for $G$ and $f(s)=g(s)$ for all $s\in S$, then $f=g$.

Indeed, the condition on $S$ tells us that, for each $x\in G$, there are $s_1,s_2,\dots,s_n\in S$ and $\varepsilon_k=\pm1$ for $k=1,2,\dots,n$ with $$ x=s_1^{\varepsilon_1}s_2^{\varepsilon_2}\dots s_n^{\varepsilon_n} $$ and so \begin{align} f(x)&=f(s_1^{\varepsilon_1}s_2^{\varepsilon_2}\dots s_n^{\varepsilon_n})\\ &=f(s_1)^{\varepsilon_1}f(s_2)^{\varepsilon_2}\dots f(s_n)^{\varepsilon_n}\\ &=g(s_1)^{\varepsilon_1}g(s_2)^{\varepsilon_2}\dots g(s_n)^{\varepsilon_n}\\ &=g(s_1^{\varepsilon_1}s_2^{\varepsilon_2}\dots s_n^{\varepsilon_n})\\ &=g(x) \end{align}

The notation on $\mathbb{Z}$ is additive, but it's just a question of notation. It's clear that $\{1\}$ is a generating set for $\mathbb{Z}$. Thus two homomorphisms $f\colon\mathbb{Z}\to S_3$ and $g\colon\mathbb{Z}\to S_3$ such that $f(1)=g(1)$ are the same homomorphism.

The group $\mathbb{Z}$ has however a very special property: if $G$ is a group, for any $x\in G$ there is a homomorphism $\varphi_x\colon\mathbb{Z}\to G$ with $\varphi_x(1)=x$. This is a stronger property than $\{1\}$ being a generating set (it's usually expressed by saying that $\mathbb{Z}$ is a free group on $\{1\}$).

So the number of homomorphisms from $\mathbb{Z}$ to $G$ is always the same as the cardinality of $G$.

$\endgroup$
  • $\begingroup$ I see. egreg, could you also explain why this is true for the group Z, and if this is true only for the group Z? Thanks for your answer. $\endgroup$ – jstnchng Oct 19 '14 at 21:00
  • $\begingroup$ @JC1397 Once you have selected $x\in G$, you can (and actually must) define $\varphi_x(n)=x^n$, for $n\in\mathbb{Z}$, because $n=\underbrace{1+1+\dots+1}_{n \text{ times}}$ for $n>0$, so necessarily $\varphi_x(n)=\underbrace{xx\dots x}_{n\text{ times}}$. Complete for $n<0$. $\endgroup$ – egreg Oct 19 '14 at 21:08
  • $\begingroup$ I see. Thanks a lot for a detailed answer. $\endgroup$ – jstnchng Oct 19 '14 at 21:33
3
$\begingroup$

It means that if we give the value of $f(1)$ then we have the value of $f(n)$ for all $n\in\Bbb Z$ since

$$f(n)=f(\underbrace{1+\cdots+1}_{n\;\text{times}})=f(1)\circ\cdots \circ f(1)$$

but since there's $6$ elements of $S_3$ then there's $6$ homomorphisms from $\Bbb Z$ to $S_3$.

$\endgroup$
2
$\begingroup$

Hint: use the property that for any homomorphism $\phi$, $\phi(xy) = \phi(x)\phi(y)$. Since $1$ generates $\mathbb{Z}$ under addition, i.e. for any positive integer $n$ we have $$n = \underbrace{1+1+\cdots+1}_{n \text{ times}}$$,

we have $$\phi(n) = \phi(\underbrace{1+1+\cdots+1}_{n \text{ times}}) = \underbrace{\phi(1) \circ \phi(1) \circ \cdots \circ \phi(1)}_{n \text{ times}}$$

Hence, as you can see, the image of $\mathbb{Z}$ under $\phi$ (that is, the set $\{\phi(z) \mid z \in \mathbb{Z}\}$ is completely determined by where $\phi$ takes $1$. Once we know what $\phi(1)$ is, we know that $\phi(n) = \phi(1)^{n}$ (or $n\phi(1)$, if you prefer additive notation), since $\phi$ is a homomorphism - there is no flexibility here.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.